Problem: Simplify; express your answer in exponential form. Assume $a\neq 0, r\neq 0$. $\dfrac{{(a^{5})^{-4}}}{{(ar^{5})^{-3}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${a^{5}}$ to the exponent ${-4}$ . Now ${5 \times -4 = -20}$ , so ${(a^{5})^{-4} = a^{-20}}$ In the denominator, we can use the distributive property of exponents. ${(ar^{5})^{-3} = (a)^{-3}(r^{5})^{-3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(a^{5})^{-4}}}{{(ar^{5})^{-3}}} = \dfrac{{a^{-20}}}{{a^{-3}r^{-15}}}$ Break up the equation by variable and simplify. $\dfrac{{a^{-20}}}{{a^{-3}r^{-15}}} = \dfrac{{a^{-20}}}{{a^{-3}}} \cdot \dfrac{{1}}{{r^{-15}}} = a^{{-20} - {(-3)}} \cdot r^{- {(-15)}} = a^{-17}r^{15}$.